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vekshin1
2 years ago
7

At what location does gravity play a role in moving tectonic plates

Physics
2 answers:
Ksju [112]2 years ago
8 0
At the edge of the plates.
Lena [83]2 years ago
7 0

Answer:

In subduction of one plate over another on the basis of weight

Explanation:

When two plates moving and colliding, the plate with higher mass will subduct below the plate with lower weigh and lighter would remain above. generally continental types of plates are lighter in weight and oceanic plates are heavier.

Ex:- The collision of Eurasian plate and Indian plate. Indian plate is below and Eurasian plate above and formation of Himalaya

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From the change of GPE into KE. Conservation of energy tells us this.
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If Earth were twice as far from the sun, the force of gravity attracting the Earth to the Sun would be
zalisa [80]
If Earth was twice as far from the sun, the force of gravity attracting the Earth to the sun would be only one-quarter as strong. The correct answer will be C. 
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Calculate the change in length of a Pyrex glass dish (Coefficient of linear expansion for Pyrex is 3 * 10-6 / oC) that is 0.25 m
DIA [1.3K]

9*10^{-5} m

Explanation:

Step 1:

We are given the initial length of the Pyrex glass dish at a particular temperature and need to calculate the change in the length when the temperature changes by 120° C. The coefficient of linear expansion of Pyrex is provided.

Step 2:

Change in length = Coefficient of linear expansion * Change in temperature * Initial length

Step 3:

Coefficient of linear expansion = 3*10^{-6} /°C

Change in temperature = 120°C = 120 K

Initial length = 0.25 m

Step 4:

Change in length = 3*10^{-6} * 120 * 0.25 = 9*10^{-5} m

8 0
3 years ago
A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the
BartSMP [9]

Answer:

r=1.14m

Explanation:

\theta is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

F_m=F_c\\qvB=F_c(1)

F_c is the centripetal force and is defined as:

F_c=m\frac{v^2}{r}

Here v is the proton's speed and r is the radius of the circular motion. Replacing this in (1) and solving for r:

qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}

Recall that 1 J is equal to 6.242*10^{12}MeV, so:

4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J

We can calculate v from the kinetic energy of the proton:

K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}

Finally, we calculate the radius of the proton path:

r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m

8 0
3 years ago
What is the momentum of and 200 kg car traveling south at 22 m/sec?
motikmotik
The momentum of the car is 4.4x10^3 kg•m/sec
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